#### Answer

$f$ and $g$ are inverses of each other

#### Work Step by Step

The two functions are inverses if $(f\circ g)(x)=x$ and if $(g\circ f)(x)=x$
First, $f(g(x))=\frac{\frac{2x+1}{x-1}+1}{\frac{2x+1}{x-1}-2}$.
After multiplying both the nominator and the denominator by (x-1), we get:
$\frac{2x+1+x-1}{2x+1-2x+2}=\frac{3x}{3}=x$.
Second, $g(f(x))=\frac{2\frac{x+1}{x-2}+1}{\frac{x+1}{x-2}-1}$.
After both the numerator and the denominator by $x-2$, we get
$\frac{2x+2+x-2}{x+1-x+2}=\frac{3x}{3}=x$
Therefore, $f$ and $g$ are inverses of each other.